Question

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A mail-order company business has six telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table. 4 5 6 0.18 0.07 0.03 X 0 2 3 p(x) 0.12 0.15 0.20 0.25 Calculate the probability of each of the following events. (a) at most three lines are in use) (b) {fewer than three lines are in use) (c) (at least three lines are in use} (d) {between two and five lines, inclusive, are in use) (e) {between two and four lines, inclusive, are not in use) {at least four lines are not in use} Need Help? Read It Watch It Talk to a Tutor

Answer 1

Fay ND.. Date: Answer: A mes oreler company business has six telephone lines. Lety denote the number of line is use at a specific time 2020/5/17 08:46 х 2 6 pex 0) 2 0.15 0.26 0.25 0.18 | 007 I. asen Now calcetate following probability atmost three lines are in use} @ PCX33) = pix=0) Aprxen fpcx=2) +P1X=3) = 0.12 +0.15+ 0.20+0.25 IPIX<3) = 0.72 { fewer than 3 line are in usey prx(3)=P(x=0) + P(x=1&P(x=2) PrX43) = 0.12 +0.15+ 0.20 TP( xc3 = 0.47 at least three lines are in use? PCX2,3) = P(X=3) TP(X=4) +PCX55) +PCX=6) = 0.25 to 18 to 0.07 +0.03 (P (X2, 3) = 0.53

Date: a {between two and five lines (both inclusive are in uses 2020/5/17 08:41 PC 25X55) = PCX 2) + pox=32 +P(X=4)+por = 0.207 0.25+0.18 +0.07 Ip12<$5)= 0.70 { between two and fourlines. (both inclusive) are in use. PC2SX34)=P(x=2) +p(x=3) + P(x=4) = 0.20+0.25+ 0.18 /pc2&X<4) = 0.63 {at least four line are not in use P(X24) = pcx0) +P081) +P(x2) TP1x=3) PIXCW = 0.12+ 0 15+0.20 +0.25 P(X24) = 0.72